Xs8, Xs9 - Second order non-homogeneous recurrences | OCR A-Level Further Mathematics B (MEI)

Combine a homogeneous family with an appropriate trial solution. Handle polynomial and exponential forcing, resolve overlap, and use complete solutions to justify limits, model validity and the behaviour of associated sequences.

The complete solution has two ingredients

Can a sequence grow without bound while the ratio of consecutive terms settles to a finite number?

Yes. A recurrence's solution and an associated sequence can have very different limits. First we need a dependable way to obtain the complete solution.

You already know

The previous lesson solved homogeneous second order recurrences using distinct, repeated or complex auxiliary roots. First order non-homogeneous solutions combined a homogeneous family with one trial solution, and initial conditions applied to their sum.

The same addition principle works here. Let $\mathcal L[u]_n=\alpha u_{n+2}+\beta u_{n+1}+\gamma u_n$ , where the coefficients are constant. This notation is only a short name for the displayed linear expression. If $\mathcal L[h]_n=0$ and $\mathcal L[p]_n=f(n)$ , then $\mathcal L[h+p]_n=f(n)$ .

Homogeneous family plus trial solution

For $\mathcal L[u]_n=f(n)$ ,

u_n=h_n+p_n,

where $h_n$ is the full two-constant homogeneous family and $p_n$ is one trial solution for the forcing. Try a constant for constant forcing and $An+B$ for linear forcing, provided the trial does not overlap the homogeneous family.

Not in the formula booklet: learn it.

Keep the equation in its supplied form. An index shift would change the forcing too; it is usually unnecessary.

Worked example 1: constant forcing

Question. Solve $u_{n+2}-5u_{n+1}+6u_n=4$ , with $u_0=6$ , $u_1=13$ .

Plan. Find the homogeneous roots, add a constant trial and then use both conditions.

\begin{aligned} r^2-5r+6&=(r-2)(r-3)=0 && \text{homogeneous equation}\\ h_n&=A2^n+B3^n && \text{two-constant family}\\ K-5K+6K&=4\implies K=2 && \text{constant trial}\\ u_n&=A2^n+B3^n+2 && \text{general solution}\\ A+B&=4,\quad2A+3B=11 && \text{apply conditions to the sum}\\ B&=3,\quad A=1 && \text{solve} \end{aligned}

Why this step? The trial contributes $2$ to both initial values; ignoring it would select the wrong constants.

Answer. $u_n=2^n+3\,3^n+2$ .

Check. The initial values are $6$ and $13$ . The geometric terms have zero residual, while the constant gives $2-10+12=4$ .

Change only the forcing from $4$ to $4n$ . Keep the homogeneous equation and both initial conditions.

Worked example 2: linear forcing

Question. Solve $v_{n+2}-5v_{n+1}+6v_n=4n$ , with $v_0=6$ , $v_1=13$ .

Plan. Try $p_n=an+b$ , then add the same homogeneous family.

\begin{aligned} p_{n+2}&=a(n+2)+b && \text{shift by two}\\ \mathcal L[p]_n&=2an+(-3a+2b) && \text{substitute and collect}\\ 2a&=4,\quad-3a+2b=0 && \text{compare coefficients}\\ a&=2,\quad b=3 && \text{trial }p_n=2n+3\\ v_n&=A2^n+B3^n+2n+3 && \text{general solution}\\ A+B&=3,\quad2A+3B=8 && \text{initial conditions}\\ B&=2,\quad A=1 && \text{solve} \end{aligned}

Why this step? Shifting $an+b$ produces a constant contribution, so the trial needs $b$ even though $4n$ has no constant term.

Answer. $v_n=2^n+2\,3^n+2n+3$ .

Check. $v_0=6$ , $v_1=13$ , and $\mathcal L[2n+3]_n=4n$ .

“General solution” means both arbitrary homogeneous constants remain. “Particular solution” may refer to a trial solving the equation or to the sequence after conditions are imposed; write the full equality so the role is clear.

Watch out

Wrong: apply the initial values to $h_n$ , then add $p_n$ afterwards. Right: apply every condition to $u_n=h_n+p_n$ . The trial contributes to the starting terms.

Find the full family for a new linear forcing.

A quadratic needs all its coefficients

A missing term in the forcing is a coefficient of zero to match, not permission to omit that term from the trial. Shift first, expand next, then collect one power at a time.

Quadratic forcing

For quadratic $f(n)$ , first try

p_n=an^2+bn+c.

If $1$ is not an auxiliary root, this full quadratic trial is sufficient. The homogeneous roots may be distinct, repeated or complex: their family is still added in the appropriate form.

Not in the formula booklet: learn it.

The following forcing is only $2n^2$ , yet a non-zero trial constant will be essential.

Worked example 3: no constant in the forcing

Question. Solve $u_{n+2}-5u_{n+1}+6u_n=2n^2$ , with $u_0=8$ , $u_1=17$ .

Plan. Use the full quadratic trial and retain the original indexing.

\begin{aligned} p_{n+2}&=an^2+(4a+b)n+(4a+2b+c) && \text{expand the shift}\\ p_{n+1}&=an^2+(2a+b)n+(a+b+c) && \text{other shift}\\ \mathcal L[p]_n&=2an^2+(-6a+2b)n+(-a-3b+2c) && \text{collect}\\ 2a&=2,\quad-6a+2b=0,\quad-a-3b+2c=0 && \text{compare}\\ a&=1,\quad b=3,\quad c=5 && \text{solve}\\ u_n&=A2^n+B3^n+n^2+3n+5 && \text{general solution}\\ A+B&=3,\quad2A+3B=8 && \text{initial values}\\ A&=1,\quad B=2 && \text{solve} \end{aligned}

Why this step? The shifts create lower powers. The trial constant $5$ cancels the unwanted constant left by its quadratic and linear terms.

Answer. $u_n=2^n+2\,3^n+n^2+3n+5$ .

Check. The polynomial residual is $2n^2$ . The full expression gives $u_0=8$ , $u_1=17$ , as required.

Keep the forcing and initial values but change the homogeneous operator to $u_{n+2}+u_n$ . Its roots are now complex; the polynomial calculation remains entirely real.

Worked example 4: adding a trigonometric family

Question. Solve $v_{n+2}+v_n=2n^2$ , with $v_0=8$ , $v_1=17$ .

Plan. Find the complex-root homogeneous family, then determine the quadratic trial.

\begin{aligned} r^2+1&=0\implies r=\pm i && \text{modulus }1,\text{ angle }\pi/2\\ h_n&=A\cos(n\pi/2)+B\sin(n\pi/2) && \text{homogeneous family}\\ p_{n+2}+p_n&=2an^2+(4a+2b)n+(4a+2b+2c) && \text{full trial}\\ 2a&=2,\quad4a+2b=0,\quad4a+2b+2c=0 && \text{compare}\\ a&=1,\quad b=-2,\quad c=0 && \text{solve}\\ v_n&=A\cos(n\pi/2)+B\sin(n\pi/2)+n^2-2n && \text{general solution}\\ A&=8,\quad B-1=17\implies B=18 && \text{initial values} \end{aligned}

Why this step? The forcing determines the trial, while the homogeneous roots determine the complementary shapes. Neither replaces the other.

Answer. $v_n=8\cos(n\pi/2)+18\sin(n\pi/2)+n^2-2n$ .

Check. The quadratic gives $(n+2)^2-2(n+2)+n^2-2n=2n^2$ ; shifting either trigonometric term by two reverses its sign, so its residual is zero.

Here the final constant vanishes, but that follows from the coefficient equations. It could not safely have been assumed in advance.

Watch out

Wrong: omit the constant because the forcing has no constant. Right: retain $an^2+bn+c$ , then compare all three coefficients. An index shift creates lower powers.

Use a complete trial with this new quadratic forcing.

When the root one gets in the way

A constant is $1^n$ . If $1$ is an auxiliary root, a constant trial already belongs to the homogeneous family and has zero residual. If $1$ is a repeated root, a linear trial also belongs to that family.

Polynomial overlap

For polynomial forcing of degree $d$ , start with a full polynomial of degree $d$ . Multiply it by $n^m$ , where $m$ is the multiplicity of auxiliary root $1$ : use $m=0$ if $1$ is absent, $m=1$ for a simple root and $m=2$ for a repeated root.

Not in the formula booklet: learn it.

Only the trial changes. Keep the full homogeneous family as well.

Worked example 5: a simple root one

Question. Solve $u_{n+2}-3u_{n+1}+2u_n=4$ , $u_0=1$ , $u_1=2$ .

Plan. The roots are $1,2$ , so try $Kn$ for the constant forcing.

\begin{aligned} h_n&=A+B2^n && \text{homogeneous family}\\ \mathcal L[Kn]_n&=K\bigl((n+2)-3(n+1)+2n\bigr) && \text{substitute}\\ &=-K=4\implies K=-4 && \text{match}\\ u_n&=A+B2^n-4n && \text{general solution}\\ A+B&=1,\quad A+2B-4=2 && \text{conditions}\\ B&=5,\quad A=-4 && \text{solve} \end{aligned}

Why this step? A constant trial would give $K-3K+2K=0$ , regardless of $K$ .

Answer. $u_n=5\,2^n-4n-4$ .

Check. The formula gives $u_0=1$ , $u_1=2$ ; the trial residual is $-(-4)=4$ .

Now make root $1$ repeated, keeping the forcing and the initial values. The required extra factor changes from $n$ to $n^2$ .

Worked example 6: a repeated root one

Question. Solve $v_{n+2}-2v_{n+1}+v_n=4$ , $v_0=1$ , $v_1=2$ .

Plan. Use $A+Bn$ for the homogeneous family and $Kn^2$ for the forcing.

\begin{aligned} h_n&=A+Bn && (r-1)^2=0\\ \mathcal L[Kn^2]_n&=K\bigl((n+2)^2-2(n+1)^2+n^2\bigr) && \text{substitute}\\ &=2K=4\implies K=2 && \text{second difference}\\ v_n&=A+Bn+2n^2 && \text{general solution}\\ A&=1,\quad A+B+2=2\implies B=-1 && \text{conditions} \end{aligned}

Why this step? Both a constant and a linear expression have zero second difference. A quadratic is the first degree that can produce a non-zero constant.

Answer. $v_n=2n^2-n+1$ .

Check. Its second difference is $4$ , and its first two terms are $1,2$ .

This explains a useful connection: constant second differences characterise quadratic sequences, just as constant first differences characterise arithmetic sequences.

Watch out

Wrong for a repeated root $1$ and non-zero constant forcing: try $Kn$ . Right: try $Kn^2$ , since every linear sequence already has zero second difference.

Keep the repeated root and change the forcing and initial values.

Match the base before the coefficient

For forcing $dk^n$ , try $Kk^n$ unless $k$ is an auxiliary root. A simple matching root needs a factor $n$ ; a repeated matching root needs $n^2$ .

Exponential forcing

For non-zero $k$ , use

p_n=Kn^mk^n,

where $m=0,1,2$ is the multiplicity of $k$ as an auxiliary root. Substitute into the original recurrence to find $K$ , then add the homogeneous family and apply conditions.

Not in the formula booklet: learn it.

A repeated-root homogeneous family already contains both $k^n$ and $nk^n$ . The trial must supply a third shape.

Worked example 7: no overlap

Question. Solve $u_{n+2}-4u_{n+1}+4u_n=3^n$ , $u_0=u_1=0$ .

Plan. The auxiliary root is $2$ , repeated, but the forcing base is $3$ , so try $K3^n$ .

\begin{aligned} h_n&=(A+Bn)2^n && \text{repeated-root family}\\ \mathcal L[K3^n]_n&=K3^n(9-12+4)=K3^n && \text{substitute}\\ K&=1 && \text{match }3^n\\ u_n&=(A+Bn)2^n+3^n && \text{general solution}\\ A+1&=0,\quad2(A+B)+3=0 && \text{conditions}\\ A&=-1,\quad B=-\tfrac12 && \text{solve} \end{aligned}

Why this step? Repeated homogeneous roots do not force an extra $n$ into every trial; overlap with the forcing base is what matters.

Answer. $u_n=3^n-(1+\tfrac12n)2^n$ .

Check. Both starting values are zero, and the base- $3$ trial has residual $3^n$ .

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Xs8, Xs9 - Second order non-homogeneous recurrences

The complete solution has two ingredients

A quadratic needs all its coefficients

When the root one gets in the way

Match the base before the coefficient

Continue this lesson.

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